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New sharp inequalities related to classical trigonometric inequalities
Abd Chouikha
To cite this version:
Abd Chouikha. New sharp inequalities related to classical trigonometric inequalities. Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales, 2021, Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A. Matemáticas, volume 115, (2021) (Article number:
143). �hal-02514447v2�
New sharp inequalities related to classical trigonometric inequalities
Abd Raouf Chouikha
*Abstract
In this paper, we establish sharp inequalities for trigonometric func- tions. We prove for 0< x < π2
cos(x) +x3 (
1−x2 63
)sin(x) 15 <
(sin(x) x
)3
< cos(x) + x3sin(x) 15 . This improves some bounds framing the function sin(x)x and generalizes some inequalities chains.
Key Words and phrases: Trigonometric functions; Inequalities.1
1 A wellknown inequalities
For 0 < x < π/2 the following inequalities are wellknown in the literature that
(cos(x))1/3 < sin(x)
x < 2 + cos(x)
3 (1)
The left-hand side is known as Adamovic-Mitrinovic inequality (see [1- 2]), while the right-hand side is known as Cusa inequality. The latter one which was proved by Huygens was used to estimate the number π, [3]. The inequalities (1) have been attracted by many people and have inspired a lot
*chouikha@math.univ-paris13.fr. 4, Cour des Quesblais 35430 Saint-Pere
12010 Mathematics Subject Classication : 26D07, 33B10, 33B20, 26D15
of interesting papers, see for example [4] and the references therein.
By using inequalities involving several means, Neuman [4] presented the fol- lowing inequality chain generalizing the Adamovic-Mitrinovic inequality. For x∈(0,π2) we have
(cos(x))1/3 <
(
cos(x)sin(x) x
)1/4
<
( sin(x) arctan(sin(x))
)1/2
(2)
<
[1
2(cos(x) + sin(x) 2x )
]1/2
<
[1 + 2 cos(x) 3
]1/2
<
[1 + cos(x) 2
]2/3
< sin(x)
x .
Thus the left inequality of (1.1) is improved.
Yang [6] proved that for 0< x < π2, sin(x)
x <
(2 3cos(x
2) + 1 3
)2
<(cos(x
3))3 < 2 + cos(x)
3 (3)
which improves the right inequality of (1).
Motivated by (1) and the dierent sharp bounds, in Sections 2 and 3 we establish nest inequalities than those known before for trigonometric functions
1 + x4
15 <1 +x3
(
1− x2 63
)tan(x)
15 < 1 cos(x)
(sin(x) x
)3
<1 + x3tan(x) 15 . By using certain estimates, we complete inequality chains (2) and (3), im- proving therefore inequality (1). In Section 4 we examine the incidence of these results on the Wilker type inequalities.
More precisely we establish the following inequality chain for 0< x < π/2 1
2
(sin(x) x
)2
+tan(x)
2x > 1 + x3tan(x)
15 > 1 cos(x)
(sin(x) x
)3
> 1 +x3
(
1− x2 63
)tan(x) 15 > 2
3
(sin(x) x
)
+ tan(x) 3x
2 The Adamovic-Mitrinovic inequality
The rst inequality in (1.1) is equivalent to x
tan(x) <
(sin(x) x
)2
; 0 < x < π 2 as well as equivalent to
1 cos(x)
(sin(x) x
)3
>1
The following result gives a lower estimate of Adamovic-Mitrinovic inequality better than those known. This allows in particular to slightly improve the chain (2)
Theorem 2-1 For0< x < π/2 the following inequalities hold cos(x) <
(
cos(x)sin(x) x
)3/4
< cos(x) + x4cos(x) 15
< cos(x) +x3
(
1− x2 63
)sin(x) 15 <
(sin(x) x
)3
(4) Moreover, there exists 0 < x0 < π/2 such that for 0 < x < x0 < π/2 the following inequalities hold
[1 + cos(x) 2
]2
<cos(x)+x4cos(x)
15 < cos(x)+x3
(
1− x2 63
)sin(x) 15 <
(sin(x) x
)3
. (5) Proof Indeed, Let us consider the following trigonometric functions with power series. We will use the Taylor expansions of sin(x), and cos(x)
cos(x) = 1− x2 2! + x4
4! − x6
6! +....+ (−1)kx2k
2k! + (−1)k+1 cosθx
(2k+ 2)!x2k+2
sin(x) = x−x3 3! + x5
5! −x7
7! +....+ (−1)k−1 x2k−1
(2k−1)! + (−1)k sinθx
(2k+ 1)!x2k+1
where 0< θ <1.
It is easy to remark that 1− x2
2! +x4 4! − x6
6! <cosx <1− x2 2! +x4
4! − x6 6! +x8
8!
1− x2 3! +x4
5! − x6
7! < sinx
x <1− x2 3! +x4
5! − x6 7! + x8
9!
for 0< x < π2.
Moreover we may assert the following 1−1
2x2+ 13
120x4− 41 3024x6 <
(sinx x
)3
<1−1
2x2+ 13
120x4− 41
3024x6+ 671 604800x8 On the other hand, thanks to Maple a calculation gives
(1−x2 2!+x4
4!−x6
6!)(1+x4
15)<cos(x)(1+x4
15)<(1−x2 2!+x4
4!−x6 6!+x8
8!)(1+x4 15)
<1−x2
2 + 13x4
120 − 5x6
144 +113x8
40320 <1− x2
2 +13x4
120 − 41x6 3024 <
(sinx x
)3
since
−5x6
144 +113x8
40320 + 41x6
3024 = −4x6
189 + 113x8
40320 = x6 120960
(−2560 + 339x2)<0 for 0< x < π/2.
Then, we have proved the right inequality of (4).
For the middle inequality it suces to write the dierence cos(x)+x3
(
1− x2 63
)sin(x)
15 −(cos(x)+x4cos(x) 15 ) = x3
15
(
(1− x2
63)(sin(x)
x )−cos(x)
)
Then we remark that (1− x2
63)(sin(x)
x )−cos(x)>(1− x2
63)(1− x2
6 )−(1− x2 2 + x4
24)
= 20
63x2− 59
1512x4 = 1
1512x2(480−59x2)>0.
Thus we prove for 0< x < π/2 cos(x) +x3
(
1− x2 63
)sin(x)
15 >cos(x) + x4cos(x) 15
For the left inequality, notice that we have seen above that 1− 1
2x2 + 13
120x4− 5
144x6 <cos(x)(1 + x4 15).
On the other hand a calculation yields
(cos(x) sin(x) x
)3/4
< 1−1
2x2+ 7 120x4. Now
1− 1
2x2+ 13
120x4− 5
144x6−(1− 1
2x2+ 7
120x4) = 1
20x4− 5
144x6 >0 for 0< x < π/2. So, (4) is proved.
To prove (5) it suces to notice that from the frame ofcos(x)we deduce 1−1
2x2+ 5
48x4− 17 1440x6 <
[1 + cos(x) 2
]2
<1− 1
2x2+ 5 48x4. Now we have
1−1
2x2+ 5
48x4 <1− 1
2x2+ 13
120x4− 5
144x6 <cos(x) + x4cos(x) 15 . as soon as x < x0 = 0.346410< π2.
This means for 0< x < x0 one has
[1 + cos(x) 2
]2
<cos(x) + x4cos(x) 15 .
Remark 2-2 We may assert that the following inequality chain holds at least for 0< x < x0 = 0.346410< π2
(cos(x))<
(
cos(x)sin(x) x
)3/4
<
( sin(x) arctan(sin(x))
)3/2
<
[1
2(cos(x) + sin(x) 2x )
]3/2
<
[1 + 2 cos(x) 3
]3/2
<
[1 + cos(x) 2
]2
< cos(x) + x4cos(x) 15 <
(sin(x) x
)3
. Thus, we enrich slightly the chain (2).
3 The Cusa-Huygens inequality
The following result gives another estimate of this inequality better than those known. This allows us in particular to extend and complete the chain (2).
Theorem 3-1 For0< x < π/2 the following inequality holds
(sin(x) x
)3
< cos(x) + cos(x)x3(tan(x))
15 = cos(x) + x3sin(x)
15 (6)
Moreover there exists for 0< x < π/2 the following inequalities hold
(sin(x) x
)3
< cos(x) + x3sin(x) 15 <
(2 3cos(x
2) + 1 3
)6
(7)
Proof Indeed, take again the Taylor expansions ofsin(x) and cos(x) cos(x) = 1− x2
2! + x4 4! − x6
6! +....+ (−1)kx2k
2k! + (−1)k+1 cosθx
(2k+ 2)!x2k+2
sin(x) = x−x3 3! + x5
5! −x7
7! +....+ (−1)k−1 x2k−1
(2k−1)! + (−1)k sinθx
(2k+ 1)!x2k+1 where 0< θ <1. We easily remark that
cosx <1− x2 2! +x4
4! − x6 6! +x8
8! − x10 10! + x12
12!
= 1− 1
2x2+ 1
24x4− 1
720x6+ 1
40320x8 − 1
3628800x10+ 1
479001600x12 1−x2
3!+x4 5!−x6
7!+x8 9!+x10
11! = 1−x2 6 + x4
120− x6
5040+ x8
362880− x10
39916800 < sinx x for 0< x < π2.
By a calculation we deduce the following (sinx
x )3 <1− x2
2 +13x4
120 −41x6
3024 + 671x8
604800− 73x10
1140480 + 597871x12 217945728000 On the other hand, thanks to Maple a calculation yields
cos (x) + 1
15x3sin (x)−(sin (x))3 x3
>1−x2 2 + x4
24− x6
720 + x8
40320 − x10
3628800 + x12
479001600 − x14 87178291200 + 1
15x3
(
x−1
6x3+ 1
120x5− 1
5040x7+ 1
362880x9− 1
39916800x11
)
−
(
1− x2
2 +13x4
120 − 41x6
3024 + 671x8
604800− 73x10
1140480 + 597871x12 217945728000
)
= 1
945x6− 1
1890x8+ 1
19800x10− 2903
1135134000x12− 733
435891456000x14 Moreover, we verify by Maple that the polynomial
1
945 − 1
1890x2+ 1
19800x4− 2903
1135134000x6− 733
435891456000x8 is non negative because it has no zeros in the interval 0< x < π/2
Then, for0< x < π/2we have cos (x) + 1
15x3sin (x)− (sin (x))3 x3 >0 So we proved Inequality (6).
Turn now to inequality (7). Notice that by the same way of calculation and thanks to Maple we obtain the following
(2 3 cos(x
2) + 1 3
)6
> 1−1
2x2+11
96x4− 553
34560x6+ 11833
7741440x8− 98851 928972800x10 Moreover, a calculation yields
cos (x) + 1
15x3sin (x) < 1−1
2x2+ 1
24x4− 1
720x6+ 1
40320x8− 1
3628800 x10
+ 1
479001600x12+ x3 15
(
x−1
6x3+ 1
120x5− 1
5040x7+ 1 362880x9
)
= 1−1
2x2+ 13
120 x4− 1
80x6+ 13
22400x8− 7
518400x10+ 89
479001600x12 Therefore
cos (x)+ 1
15x3sin (x)−(2 3 cos(x
2) + 1 3
)6
< 1−1
2x2+ 13
120x4− 1
80x6+ 13 22400x8
− 7
518400x10+ 89
479001600x12−(1−x2
2 +11x4
96 −553x6
34560+11833x8
7741440− 98851x10 928972800)
=− 1
160x4+ 121
34560x6− 5243
5529600x8+ 28769
309657600x10+ 89
479001600x12 We verify by Maple that the polynomial
− 1
160 + 121
34560x2− 5243
5529600x4+ 28769
309657600x6+ 89
479001600x8 is non negative because it has no zeros in the interval 0< x < π/2 .
This means for0< x < π2 one has cos(x) + x3sin(x)
15 <
(2 3cos(x
2) + 1 3
)6
.
So inequality (7) is then proved.
Remark 3-2 (i) We were forced to consider a polynomial at least of order 12 in order to estimate trigonometric functions. Indeed, for a lower order estimate we would have an increasing polynomial which changes sign since it admits a root in the interval [0, π/2].
(ii) We may assert that the following inequality chain holds at least for 0< x < π2
(sin(x) x
)3
< cos(x) + cos(x)(tan(x))4 15 <
(2 3cos(x
2) + 1 3
)6
<
(
cos(x 3)
)9
<
(2 + cos(x) 3
)3
. Thus, we expand the chain (3).
4 A wilker type inequalities
The Wilker inequality asserts that for 0< x < π/2
(sin(x) x
)2
+ tan(x)
x >2 (8)
This inequality has been proved by [7].
The Wilker-type inequalities have been attracted by many people and have motivated a large number of research papers involving various generalizations and improvements, see [8] for example.
A related inequality no less interesting is Huygens inequality asserts that for 0<|x|< π/2
2sin(x)
x + tan(x)
x >3 (9)
Chen and Sandor [4] proved the following inequality chain for 0<|x|< π/2
1 2
(sin(x) x
)2
+tan(x)
2x > 1 cos(x)
(sin(x) x
)3
> 2 3
(sin(x) x
)
+ tan(x) 3x >
(sin(x) x
)2
3 (
tan(x) x
)1
3
> 1 2
( x sin(x)
)2
+ x
2 tan(x) > 2 3
( x sin(x)
)
+ x
3 tan(x) >1 (10) We propose the following
Theorem 4-1 For0< x < π/2 the following inequalities holds 1
2
(sin(x) x
)2
+tan(x)
2x > 1 +x3(tan(x))
15 > 1 cos(x)
(sin(x) x
)3
(11)
Proof Consider the left inequality and we write dierence after mul- tiplying by cos(x) one gets
1 2
cos (x) (sin (x))2
x2 +1
2
sin (x)
x −cos (x)− 1
15x3sin (x)
Notice that by simple calculations 1−(1
2)x2+ ( 1
24)x4−( 1
720)x6 <cos(x) x−(1
6)x3+ ( 1
120)x5−( 1
5040)x7 <sin(x) cos(x)(sin(x))2
2x2 > 1 2 − 5
12x2+ 91
720x4− 41
2016x6+ 7381 3628800x8 Therefore
1 2
cos (x) (sin (x))2
x2 +1
2
sin (x)
x −cos (x)− 1
15x3sin (x)> 1 2−5x2
12 +91x4
720 −41x6 2016 + 7381x8
3628800 + 1 2− x2
12+ x4
240 − x6
10080+ x8
725760−1 + x2 2 − x4
24+ x6 720
−x3 15
(
x− x3 6 + x5
120 − x7 5040
)
= 4x4
45 − 2x6
105 + 1231x8 604800− x3
15
(
x− x3 6 + x5
120 − x7 5040
)
= x4 45− x6
126 + 179x8
120960 + x10 75600 We verify by Maple that the polynomial
x4 45 − x6
126 + 179x8
120960+ x10
75600 = x4(13440−4800x2+ 895x4+ 8x6) 604800
is non negative because it has no zeros in the interval 0< x < π/2 . This means for0< x < π2 one has
1 2
(sin(x) x
)2
+tan(x)
2x > 1 +x3(tan(x)) 15 . So the left inequality (11) is then proved.
The right inequality has been already proved by Theorem 3-1.
Concerning the Huygens inequality we propose to prove the following
Theorem 4-2 For0< x < π/2 the following inequalities holds 1
cos(x)
(sin(x) x
)3
> 1+x3
(
1− x2 63
)tan(x) 15 > 2
3
(sin(x) x
)
+tan(x) 3x . (12) Proof Consider the right inequality we multiply the dierence bycos(x). Then we have
cos(x) +x3
(
1− x2 63
)sin(x) 15 > 2
3
(sin(x) cos(x) x
)
+sin(x) 3x . Note as we saw above the framing of trigonometric functions, then we get
cos(x)>1− 1
2x2+ 1
24x4− 1 720x6 sin(x)> x− 1
6x3+ 1
120x5− 1 5040x7 Therefore after simplication
cos(x) +x3
(
1− x2 63
)sin(x)
15 > 1− 1
2x2+ 13
120x4− 41 3024x6 A simple calculation yields
2 3
sin (x) cos (x)
x + sin (x)
3x < 1−1
2x2+ 11
120x4− 43 5040x6 Finally for 0< x < π/2 one has
cos(x) +x3
(
1− x2 63
)sin(x) 15 − 2
3
(sin(x) cos(x) x
)
+sin(x) 3x > x4
60− 19x6 3780
= 1
3780x4(63−19x2)>0
The left inequality has been already proved by Theorem 3-2.
Remark (4-3) We may deduce from Theorems 4-1 and 4-2 the fol- lowing inequality chain for 0< x < π/2
1 2
(sin(x) x
)2
+ tan(x)
2x > 1 +x3(tan(x))
15 > 1 cos(x)
(sin(x) x
)3
> 1 +x3
(
1− x2 63
)tan(x) 15 > 2
3
(sin(x) x
)
+ tan(x) . 3x
Thus, we have completed the inequality chain (10).
REFERENCES
[1] D.S. Mitrinovic and D.D. Adamovic, Sur une inegalite elementaire ou interviennent des fonc- tions trigonometriques, Univ u Beogradu. Publik.
Elektr. Fakulteta. S. Matematika i Fizika 149 (1965), 2334.
[2] D. S. Mitrinovic, Analytic inequalities, Springer-Verlag, Berlin, 1970.
[3] J. Sandor and M. Bencze, On Huygens' trigonometric inequality, RGMIA Res. Rep. Collection 8 (2005), no. 3, Article 14.
[4] C.-P. Chen and J. Sandor, Inequality chains for Wilker, Huygens and Lazarevic type inequalities, J. Math. Inequal. 8 (2014), no. 1, 5567.
[5] E. Neuman, Renements and generalizations of certain inequalities involving trigonometric and hyperbolic functions, Adv. Inequal. Appl. 1 (2012), no. 1, 1-11.
[6] Z.-H. Yang, Three families of two-parameter means constructed by trigonometric functions, J. Inequal. Appl. 2013 (2013). Article 541.
[7] J.S. Sumner, A.A. Jagers, M. Vowe and J. Anglesio, Inequalities in- volving trigonometric functions, Amer. Math. Monthly 98 (1991), 264-267.
[8] C.-P. Chen and W.-S. Cheung, Sharpness of Wilker and Huygens type inequalities, J. Inequal. Appl. 2012 (2012), 72.
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